This question has been asked several times over on mathoverflow, and I believe also on this site. Someone who is better than I at searching should be able to find the past answers.
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Pete L. ClarkMay 6 '12 at 22:15

Has the version with "real analytic" replaced with "real entire" also been asked?
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Ricky DemerMay 6 '12 at 22:16

2

If by "real entire" you mean "entire, and real on $\mathbb R$", then the interpolation proof provides such a function.
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Robert IsraelMay 6 '12 at 22:23

Well, I would tend to define "real entire" as "expressible (on $\mathbf{R}$) as a globally convergent power series with real coefficients", but I do know that that is equivalent to what you gave.
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Ricky DemerMay 6 '12 at 22:27

3 Answers
3

No. Only if you require $g$ or its coefficients to be computable. Suppose there is such an $f$, then we could just pick the points $(n,(1+\sup\{ f(z))|n-1<z<n+1\}))$, for $n=1,2,3\ldots$
and interpolate.

This was published in the American Journal of mathematics, vol. 14, p. 214. Hardy presents a proof due to Borel, in Leçons sur les séries à termes positifs, p.27:

We may replace $\phi$ with an increasing function $\Phi$ that is always positive, is pointwise larger than $\phi$, and tends to infinity, and proceed to define $f$ and show that $f/\Phi\to\infty$. Take an increasing sequence of numbers $a_n\to\infty$, and another sequence $b_n$ with
$$ a_1<b_2<a_2<b_3<a_3<\dots, $$
and define
$$ f(x)=\sum_{n\ge 1}\left(\frac x{b_n}\right)^{\nu_n}, $$
where the positive integers $\nu_n$ are strictly increasing, and satisfy $\displaystyle \left(\frac{a_n}{b_n}\right)^{\nu_n}>\Phi^2(a_n)$. Then $f$ is entire and satisfies the required property.

In detail: The series converges because, given any positive $x$, the $n$-th root of the $n$-th term is at most $x/b_n\to 0$. If $x\in[a_n,a_{n+1})$, then $f(x)>(a_n/b_n)^{\nu_n}$, so
$$ f(x)>\Phi^2(a_{n+1})>\Phi^2(x). $$
It follows that $f/\Phi^2\ge 1$ for $x\ge a_1$, and since $\Phi(x)\to\infty$, then also $f/\Phi\to\infty$, as wanted.

Hardy mentions this while discussing a result of du Bois-Reymond: Given functions $f,g\to\infty$, positive, and increasing, write $f\succ g$ iff $f/g\to\infty$.

Theorem (du Bois-Reymond). Given any "ascending scale" $(f_n)_{n\in\mathbb N}$, that is, a sequence of functions $f_n:\mathbb R\to\mathbb R$, all positive and increasing to infinity, and such that $f_1\prec f_2\prec f_3\dots$, there is a function $f$ that increases faster than any function in the scale, that is, such that $f\succ f_n$ for all $n$.

This result was generalized by several authors, beginning with Hadamard, and eventually led to Hausdorff work on what we now call Hausdorff gaps.